3.7.14 \(\int \frac {\sqrt {d+e x}}{a-c x^2} \, dx\) [614]

Optimal. Leaf size=134 \[ -\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} c^{3/4}}+\frac {\sqrt {\sqrt {c} d+\sqrt {a} e} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{\sqrt {a} c^{3/4}} \]

[Out]

-arctanh(c^(1/4)*(e*x+d)^(1/2)/(-e*a^(1/2)+d*c^(1/2))^(1/2))*(-e*a^(1/2)+d*c^(1/2))^(1/2)/c^(3/4)/a^(1/2)+arct
anh(c^(1/4)*(e*x+d)^(1/2)/(e*a^(1/2)+d*c^(1/2))^(1/2))*(e*a^(1/2)+d*c^(1/2))^(1/2)/c^(3/4)/a^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {714, 1144, 214} \begin {gather*} \frac {\sqrt {\sqrt {a} e+\sqrt {c} d} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{\sqrt {a} c^{3/4}}-\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} c^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]/(a - c*x^2),x]

[Out]

-((Sqrt[Sqrt[c]*d - Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(Sqrt[a]*c^(3/4))
) + (Sqrt[Sqrt[c]*d + Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(Sqrt[a]*c^(3/4
))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 714

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2 + a*e^2 - 2*c*d
*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1144

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2/2)*(b/q + 1), Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2/2)*(b/q - 1), Int[(d*x)^(m - 2)/(b
/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x}}{a-c x^2} \, dx &=(2 e) \text {Subst}\left (\int \frac {x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt {d+e x}\right )\\ &=-\left (\left (\frac {\sqrt {c} d}{\sqrt {a}}-e\right ) \text {Subst}\left (\int \frac {1}{c d-\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )\right )+\left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) \text {Subst}\left (\int \frac {1}{c d+\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )\\ &=-\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} c^{3/4}}+\frac {\sqrt {\sqrt {c} d+\sqrt {a} e} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{\sqrt {a} c^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 156, normalized size = 1.16 \begin {gather*} \frac {-\sqrt {-c d-\sqrt {a} \sqrt {c} e} \tan ^{-1}\left (\frac {\sqrt {-c d-\sqrt {a} \sqrt {c} e} \sqrt {d+e x}}{\sqrt {c} d+\sqrt {a} e}\right )+\sqrt {-c d+\sqrt {a} \sqrt {c} e} \tan ^{-1}\left (\frac {\sqrt {-c d+\sqrt {a} \sqrt {c} e} \sqrt {d+e x}}{\sqrt {c} d-\sqrt {a} e}\right )}{\sqrt {a} c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d + e*x]/(a - c*x^2),x]

[Out]

(-(Sqrt[-(c*d) - Sqrt[a]*Sqrt[c]*e]*ArcTan[(Sqrt[-(c*d) - Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d + Sqrt[
a]*e)]) + Sqrt[-(c*d) + Sqrt[a]*Sqrt[c]*e]*ArcTan[(Sqrt[-(c*d) + Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d
- Sqrt[a]*e)])/(Sqrt[a]*c)

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Maple [A]
time = 0.50, size = 143, normalized size = 1.07

method	result	size
derivativedivides	\(-2 e c \left (-\frac {\left (c d +\sqrt {a c \,e^{2}}\right ) \arctanh \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 c \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {\left (-c d +\sqrt {a c \,e^{2}}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 c \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )\)	\(143\)
default	\(-2 e c \left (-\frac {\left (c d +\sqrt {a c \,e^{2}}\right ) \arctanh \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 c \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {\left (-c d +\sqrt {a c \,e^{2}}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 c \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)/(-c*x^2+a),x,method=_RETURNVERBOSE)

[Out]

-2*e*c*(-1/2*(c*d+(a*c*e^2)^(1/2))/c/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)/(
(c*d+(a*c*e^2)^(1/2))*c)^(1/2))+1/2*(-c*d+(a*c*e^2)^(1/2))/c/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*
arctan(c*(e*x+d)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(-c*x^2+a),x, algorithm="maxima")

[Out]

-integrate(sqrt(x*e + d)/(c*x^2 - a), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (100) = 200\).
time = 1.03, size = 339, normalized size = 2.53 \begin {gather*} \frac {1}{2} \, \sqrt {\frac {a c \sqrt {\frac {1}{a c^{3}}} e + d}{a c}} \log \left (a c^{2} \sqrt {\frac {a c \sqrt {\frac {1}{a c^{3}}} e + d}{a c}} \sqrt {\frac {1}{a c^{3}}} e + \sqrt {x e + d} e\right ) - \frac {1}{2} \, \sqrt {\frac {a c \sqrt {\frac {1}{a c^{3}}} e + d}{a c}} \log \left (-a c^{2} \sqrt {\frac {a c \sqrt {\frac {1}{a c^{3}}} e + d}{a c}} \sqrt {\frac {1}{a c^{3}}} e + \sqrt {x e + d} e\right ) - \frac {1}{2} \, \sqrt {-\frac {a c \sqrt {\frac {1}{a c^{3}}} e - d}{a c}} \log \left (a c^{2} \sqrt {-\frac {a c \sqrt {\frac {1}{a c^{3}}} e - d}{a c}} \sqrt {\frac {1}{a c^{3}}} e + \sqrt {x e + d} e\right ) + \frac {1}{2} \, \sqrt {-\frac {a c \sqrt {\frac {1}{a c^{3}}} e - d}{a c}} \log \left (-a c^{2} \sqrt {-\frac {a c \sqrt {\frac {1}{a c^{3}}} e - d}{a c}} \sqrt {\frac {1}{a c^{3}}} e + \sqrt {x e + d} e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(-c*x^2+a),x, algorithm="fricas")

[Out]

1/2*sqrt((a*c*sqrt(1/(a*c^3))*e + d)/(a*c))*log(a*c^2*sqrt((a*c*sqrt(1/(a*c^3))*e + d)/(a*c))*sqrt(1/(a*c^3))*
e + sqrt(x*e + d)*e) - 1/2*sqrt((a*c*sqrt(1/(a*c^3))*e + d)/(a*c))*log(-a*c^2*sqrt((a*c*sqrt(1/(a*c^3))*e + d)
/(a*c))*sqrt(1/(a*c^3))*e + sqrt(x*e + d)*e) - 1/2*sqrt(-(a*c*sqrt(1/(a*c^3))*e - d)/(a*c))*log(a*c^2*sqrt(-(a
*c*sqrt(1/(a*c^3))*e - d)/(a*c))*sqrt(1/(a*c^3))*e + sqrt(x*e + d)*e) + 1/2*sqrt(-(a*c*sqrt(1/(a*c^3))*e - d)/
(a*c))*log(-a*c^2*sqrt(-(a*c*sqrt(1/(a*c^3))*e - d)/(a*c))*sqrt(1/(a*c^3))*e + sqrt(x*e + d)*e)

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Sympy [A]
time = 3.08, size = 76, normalized size = 0.57 \begin {gather*} - 2 e \operatorname {RootSum} {\left (256 t^{4} a^{2} c^{3} e^{4} - 32 t^{2} a c^{2} d e^{2} - a e^{2} + c d^{2}, \left ( t \mapsto t \log {\left (- 64 t^{3} a c^{2} e^{2} + 4 t c d + \sqrt {d + e x} \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(-c*x**2+a),x)

[Out]

-2*e*RootSum(256*_t**4*a**2*c**3*e**4 - 32*_t**2*a*c**2*d*e**2 - a*e**2 + c*d**2, Lambda(_t, _t*log(-64*_t**3*
a*c**2*e**2 + 4*_t*c*d + sqrt(d + e*x))))

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Giac [A]
time = 2.51, size = 153, normalized size = 1.14 \begin {gather*} \frac {\sqrt {-c^{2} d - \sqrt {a c} c e} {\left | c \right |} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d + \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {a c} c^{2}} - \frac {\sqrt {-c^{2} d + \sqrt {a c} c e} {\left | c \right |} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d - \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {a c} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(-c*x^2+a),x, algorithm="giac")

[Out]

sqrt(-c^2*d - sqrt(a*c)*c*e)*abs(c)*arctan(sqrt(x*e + d)/sqrt(-(c*d + sqrt(c^2*d^2 - (c*d^2 - a*e^2)*c))/c))/(
sqrt(a*c)*c^2) - sqrt(-c^2*d + sqrt(a*c)*c*e)*abs(c)*arctan(sqrt(x*e + d)/sqrt(-(c*d - sqrt(c^2*d^2 - (c*d^2 -
a*e^2)*c))/c))/(sqrt(a*c)*c^2)

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Mupad [B]
time = 0.34, size = 302, normalized size = 2.25 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {2\,\left (\left (16\,c^3\,d^2\,e^2+16\,a\,c^2\,e^4\right )\,\sqrt {d+e\,x}-\frac {16\,c\,d\,e^2\,\left (e\,\sqrt {a^3\,c^3}+a\,c^2\,d\right )\,\sqrt {d+e\,x}}{a}\right )\,\sqrt {\frac {e\,\sqrt {a^3\,c^3}+a\,c^2\,d}{4\,a^2\,c^3}}}{16\,c^2\,d^2\,e^3-16\,a\,c\,e^5}\right )\,\sqrt {\frac {e\,\sqrt {a^3\,c^3}+a\,c^2\,d}{4\,a^2\,c^3}}-2\,\mathrm {atanh}\left (\frac {2\,\left (\left (16\,c^3\,d^2\,e^2+16\,a\,c^2\,e^4\right )\,\sqrt {d+e\,x}+\frac {16\,c\,d\,e^2\,\left (e\,\sqrt {a^3\,c^3}-a\,c^2\,d\right )\,\sqrt {d+e\,x}}{a}\right )\,\sqrt {-\frac {e\,\sqrt {a^3\,c^3}-a\,c^2\,d}{4\,a^2\,c^3}}}{16\,c^2\,d^2\,e^3-16\,a\,c\,e^5}\right )\,\sqrt {-\frac {e\,\sqrt {a^3\,c^3}-a\,c^2\,d}{4\,a^2\,c^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(1/2)/(a - c*x^2),x)

[Out]

- 2*atanh((2*((16*a*c^2*e^4 + 16*c^3*d^2*e^2)*(d + e*x)^(1/2) - (16*c*d*e^2*(e*(a^3*c^3)^(1/2) + a*c^2*d)*(d +
e*x)^(1/2))/a)*((e*(a^3*c^3)^(1/2) + a*c^2*d)/(4*a^2*c^3))^(1/2))/(16*c^2*d^2*e^3 - 16*a*c*e^5))*((e*(a^3*c^3
)^(1/2) + a*c^2*d)/(4*a^2*c^3))^(1/2) - 2*atanh((2*((16*a*c^2*e^4 + 16*c^3*d^2*e^2)*(d + e*x)^(1/2) + (16*c*d*
e^2*(e*(a^3*c^3)^(1/2) - a*c^2*d)*(d + e*x)^(1/2))/a)*(-(e*(a^3*c^3)^(1/2) - a*c^2*d)/(4*a^2*c^3))^(1/2))/(16*
c^2*d^2*e^3 - 16*a*c*e^5))*(-(e*(a^3*c^3)^(1/2) - a*c^2*d)/(4*a^2*c^3))^(1/2)

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